Spectral Graph Theory and its Applications

Transcription

1 Spectral Graph Theory and its Applications Yi-Hsuan Lin Abstract This notes were given in a series of lectures by Prof Fan Chung in National Taiwan University Introduction Basic notations Let G = (V, E) be a graph, where V is a vertex set and E is an edge set (Graph ) We denote the edge set E = {{a, b}, {b, c}, } Definition We say that {u, v} E Moreover, we let if and only if u v, if and only if u is adjacent to v, if and only if u is a neighborhood of v N(v) := {x V x v}, d v := degree of v = N(v) (d v ) v V denotes the degree of sequence Example In graph, we say (d v ) v V = {d a, d b, d c, d d } = {3,,, } Theorem 3 (Erdos-Gallai, 960) Let (a i ) i=,,,n be a degree of sequence of a graph a i a i+ if and only if r a i r(r ) + i= n i=r+ min{r, a i }

2 Proof ( ) Trivial, since when the graph has r vertices, it will be less than r(r ); outside r-graph part, it will be less than the second part of the inequality ( ) Exercise! There are many points of view of the graph theory The graph theory will involve topology, algebra, analysis and probability It also can be applied to the computer science (algorithm, complexity), network science (big data) and optimization Classification There are many types of graphs We give brief definitions in the following Definition 4 We call G to be an undirected graph if E = {unordered pair} We call G is a simple graph if G satisfies undirected, no multiple angles, 3no loops The directed graph will be denoted by The hypergraph: E = {subsets} The weighted graph: w(e) R + {0} Let G = (V, E), A be an adjacency matrix, which means { if u v, A(u, v) = 0 otherwise

3 It is not hard to see d u = v A(u, v) Path u = u 0, u,, u t = v, where u i u i+ We call it to be a simple path if all u is are distinct Denote A t (u, v) =# of paths from u to v of length t Exercise5 # of k paths in G = (V, E) nd k, where n = V, d =average u degree= d u For instance, when k =, -paths, by using the Cauchyn Schwarz inequality, we have We also have d i n( d i n ) where = (,,, ) # of paths of length t in G =, A t, Exercise 6 Let ρ 0 =spectral radius of A (the largest eigenvalue of A) Prove that (, A k ) k ρ 0 = lim k, 3 Random walk on G In this section, we consider G to be undirected graphs Let P (u, v) =the probability of matrix to u from v (transition probability matrix), if u v P (u, v) = d u = A(u, v) = (D A)(u, v) 0, otherwise d u Note that P = D A with D being a diagonal degree matrix, D(u, v) = d v (not symmetric in general) Let f : V R + {0} be a probability distribution and v f(v) = {, if v = u Example 7 f 0 = χ u, where χ u (v) = 0, otherwise There are two versions to describe the graph In the previous example, after one step the probability becomes χ u (z)p (z, v) = (χ u, P )(v) v After t steps, the probability from u reaching v is χ u P t (v) Here we use the row vector multiplying with the matrix Example 8 Uniform distribution Stationary distribution ( n, n,, n ) = n Does lim t fp t exists? unique? If the existence holds, then we have lim fp t = Π, t where Π is the stationary distribution (ergodic) 3

4 Example 9 G is k-regular, d u probability matrix = k and n P =, where P is the n Example 0 Let G = G G be disjoint union, where G is k -regular, G is k -regular Then the limit Π = Π Π, where are stationary Example S k on k + vertices Π = ( k,, k, 0,, 0), Π = (0, 0,, 0, k,, k ), χ c P odd (c) = 0 and χ c P even (c) = (k,,, ) = Π is stationary k Let G = (V, E) be a graph without isolated vertices, S V We define VolS = d v v S and Moreover, Π(u) = ( ) du u V (ΠP )(v) = u v Π(u)P (u, v) = u v d u d u = When the graph is directed, we can define ( ) d Π + + = u and Π = where = u d+ u = v d v d v = Π(v) ( ) d u, Π + P = Π 4

5 Problem Can we find a function f such that fp = f? In general, there is no close form for this f 4 The note of convergence Let M be a symmetric matrix, then all its eigenvalues are real Let M t = U Λ t U, where Λ is the diagonal matrix Recall that P = D A = D (D AD )D and let M = D AD, then P t = D M t D Let φ is are orthonormal eigenvectors associated with eigenvalues ρ i of M, then φ 0 = D We will see more details in the following lecture Spectral graph theory - An introduction The tool is based on the generalized fast Fourier transformation (GFFT) Let M denote all positive entries matrices, A be adjacent matrices, A(u, v) 0, u, v Recall that the Perron-Frobenius theorem states that there exists a unique eigenvector with positive components All negative eigenvalues matrix can be considered as a Laplacian The combinatorial Laplacian Let G = (V, E) be a simple graph without isolated vertices, no loops, V = n Denote A to be adjacency matrix, D to be diagonal matrix and P = D A to be the transition probability matrix with eigenvalues ρ i = λ i, where λ i s will be introduced later Definition Let L be (normalized) Laplacian, L = I D AD L has eigenvalues λ 0 λ λ n ( V = n), if u = v, L(u, v) =, if u v, du d v 0, otherwise P (u, v) = Prob{moving to v from u}, fp (v) = u v f(u)p (u, v), where f is a row vector Let B(u, e) be oriented all edges arbitrarily, u V, e E such that, if u is the head of e, B(u, e) =, if u is the tail of e, 0, otherwise Fact B B = P A = L: combinatorial Laplacian Fact 3 We have Lf(u) = d u f(u) v,v u f(v) = (f(u) f(v)) v,v u Fact 4 L = D LD = D B B D = (D B) (D B) 5

6 Fact 5 All eigenvalues λ i 0 and λ 0 = 0 A homological view: We can regard C as -chains and C 0 as 0-chains Then B : C 0 C and B : C C 0 Let e = {u, v} be oriented edge, then B sends e to u v and B : α e e β v v e E Proposition 6 Consider the Rayleigh quotient and let f = gd, v V glg gg = gd LD g gg = flf fdf, where flf = f(u) (f(v) f(u)) u v,v u = f(u) (f(u) f(v)) u v,v u = (f(u)(f(u) f(v)) + f(u)(f(v) f(u)) {u,v} E = (f(u) f(v)) {u,v} E and fdf = v f(v) d v Therefore, the Rayleigh quotient is glg {u,v} E gg = (f(u) f(v)) x V f (x)d x For zero eigenvalue, f(u) = f(v) if u v Fact 7 φ 0 = f 0 D = D, where = v d v Fact 8 G has k connected components if and only if the zero eigenvalue has multiplicity k Proof ( ) Easy ( ) Exercise Fact 9 λ i for all i Proof By using the Rayleigh quotient, we have λ i sup f since (a b) (a + b ) {u,v} E (f(u) f(v)) x V f = sup (x)d x f {u,v} E (f(u) f(v)) u v (f (u) + f, (v)) 6

7 0 Example 0 Let A =, L = 0 θ be the n-th root of θ n =, then θ L = ( n (θ + θ + + θ n )) θ n { 0, if θ =, = n, if θ n Moreover, 0 has multiplicity and n n n n θ θ n has multiplicity n Example More complicated, we consider matrices A = 0 and L = Similar calculation, we have θ L = ( θ θ ) θ n where θ = e πij n θ θ n Let πj = ( cos n ) θ θ n Example A hypercubes Q n has n vertices, n n edges V = (0, ) n ={string of 0, of length n }={subsets of an n-set ({,, ), n}} Its eigenvalues of L are k n, k = 0,,, n with multiplicity and eigenvectors θ n k s, s [n] = {,,, n} θ s (T ) = ( ) s n More facts n and Lθ s = k n θ s, with k = s (exercise) Fact 3 G is a vertex disjoint union of G and G, then we have Λ G = Λ G Λ G, where Λ S denotes the spectrum of S Fact 4 If G is connected and G K n, then λ, 7

8 Proof Let φ 0 be the first eigenfunction with respect to λ 0 and f = D g By using the Rayleigh quotient, we have λ = inf g φ 0 g, Lg g, g = = inf f(x)dx=0 inf f, (D A)f f(x)dx=0 f, Df x u (f(x) f(u)) x f = inf (x)d R(f), x f(x)dx=0 where R(f) is the standard Rayleigh quotient Now, we choose f by picking different vertices a and b with a b, d b, if v = a, f (v) = d a, if v = b, 0, otherwise Then we have λ R(f) = d ad b + d bd a d a d b + d bd = a Fact 5 Suppose G has diameter D, then λ v V d v, where = D() Proof Suppose that λ = R(f) for some f, f is the harmonic eigenfunction f(x)dx = 0, and choose u 0 such that f(u 0 ) = max u f(u), then v such that f(u 0 )f(v) < 0 and a path P = {u 0, u,, u t = v}, t D Again, x u (f(x) f(u)) λ = x f ( {x,y} E(P ) f(x) f(y) ) (x)d x () f(u 0 ( x,y E(P ) f(x) f(y) ) D() f(u 0 ) f(u 0) f(u t ) D() f(u 0 ) D() Remark 6 λ n 3 Exercise 7 Find connected graphs with λ being the minimum 3 More results for discrete Laplacian 3 Cartesian product Let G i = (V i, E i ) be graphs G G has vertex set V V Definition 3 We say (u, u ) (v, v ) if and only if u = v and u v or u = v and u v By the definition, it is easy to see Q n = P P P Let G = (V, E) be a graph with V = n and L = I D AD Recall the eigenvalues are 0 = λ 0 λ λ n, if λ 0 = 0, then the graph G has disconnected components In further, the spectral gap λ λ 0 plays an important role in the spectral graph theory 8

9 3 Vertex transitive and edge transitive Let G = (V, E) be a graph and σ be (graph) automorphism such that σ : V V satisfying x y if and only if σ(x) σ(y) Definition 3 We say that G is vertex transitive if u, v V, σ automorphism such that σ(u) = v We call G to be edge transitive if e, e E, σ automorphism such that σ(e) = e Exercise 33 Find examples of graphs such that vertex transitive will not imply the edge transitive Simultaneously, the edge transitive also cannot imply the vertex transitive Hint: We can consider the Cartesian product of two triangles Note that in the following graph, the vertices are transitive, but the edges are not Theorem 34 If G is edge transitive, then λ D Theorem 35 If G is vertex transitive and G has degree k, then λ kd Definition 36 We define the index of G to be IndG = min i E i, where E i is the i-th equivalent class of edges (the graph above also shows that these two triangular have equivalent edges) Theorem 37 λ D indg if G is vertex transitive Proof By using the Rayleigh quotient again, if we choose c = λ = inf f = n k inf f λ = inf sup f c f(x)dx, then we have dx x u (f(x) f(u)) x f (x)d x ( f(x)d x) dx x u u,v (f(x) f(u)) (f(u) f(v)) x u (f(x) f(u)) x (f(x), c) d x = inf f x u (f(x) f(u)) u,v (f(u) f(v))d ud v 9

10 Fix a vertex x 0, y V, we consider P (x 0, y) of length D and P (x 0 ) = {P (x 0, y) y V } For x V, x = σ(x 0 ), P (x) = {σp (x 0, y)} Now, for every edges e E, we set N e = # of recurrence of e in P (x) In addition, N e n D E i n D min E i n DindG Let e = {u, v} and f(e) = f(u) f(v), then (f(x) f(y)) ( f(e)) x y x y e P (x,y) x Therefore, we can conclude λ e E f (e)d N e e E ndindg k D y e P (x,y) f (e)d ndindg k n e E f (e) k u,v (f(u) f(v)) D indg f (e) Let G = (V, E) undirected weighted graph Recall that we have introduced P (x, y) = Prob{moving to y from x}, 0 P (x, y) and y P (x, y) = Let Π be the stationary distribution, then we have ΠP = Π, which means Π(x)P (x, y) = ΠP (y) x Definition 38 We call P (x, y) to be the reversible random walk if which means P (x, y) is reversible Π(x)P (x, y) = Π(y)P (y, x), Let ω xy = cπ(x)p (x, y), then ω xy = ω yx and Π(x)P (x, y) = c ω xy = d x ω xy d x = Π(y)P (x, y), where d x = y ω xy The typical random walk is P (x, y) = ω xy, Π(x) = d x d x and = x G d x Recall that in Section, we have proposed the following questions: Does fp t converge? How long does it take? How about the rates of convergence? Here are partial answers The necessary (also sufficient) conditions for the convergence is Irreducibility - This is equivalent to the connectivity of graphs Aperiodicity - This is equivalent to non-bipartite of graphs If P = D A = D (D AD )D = D MD, we have P t = D M t D Let φ i be orthonormal eigenvectors for M, i = 0,, n, if we write fd = 0

11 αi φ i, α i = fd (we denote L ) D = f = and φ 0 = D Thus, fp t Π = fd M t D Π = fd (I L) t D Π n = α i ( λ i ) t φ i D i=0 (max i>0 λ i t )C 0 as t, Π = α i ( λ i ) t φ i D if λ := λ i <, for some constant C > 0 Moreover, we have the following main results λ > 0 if and only if the graph is connected if and only if irreducibility holds λ n < if and only if the graph is non-bipartite if and only if the aperiodicity holds 33 Lazy walk We give examples of the lazy walk in the following Example 39 Z = I + P i>0, λ = λ, adding loops ω u,u = d u Example 30 Z = ci + P c +, choose the best constant c (eg c = λ + λ n ) 4 Cheeger s inequality - graph version Recall that L has eigenvalues 0 = λ 0 λ λ n, where λ controls the connectivity of graphs Moreover, λ gives information about the convergence of random walks In the spectral graph theory, we consider two important problems: Expansion property and isoperimetric property 4 Isoperimetric property In the graph theory, there are two kinds of boundaries Definition 4 We call δ(s) to be a vertex boundary if δ(s) = {v V v / S, v u for some u S} We call (S) to be an edge boundary if (S) = {{u, v} E u S, v / S} For a specific volume (say 00), what is the subset with volume 00 minimizing the vertex/edge boundary? Let G = (V, E) be a graph with V = n, we want to discuss the Cheeger s inequality to help us to us to understand the isoperimetric inequality

12 4 Cheeger s inequality Recall that Vol(S) = v S d v Definition 4 (Cheeger s ratio) h(s) = S Vol(S) Definition 43 The Cheeger constant h G of G is h G := min{h(s) S V, Vol(S) Vol(G)} Here we only consider all graphs with finite edges and vertices It is easy to see that for any S with Vol(S) Vol(G), S h GVol(S) Theorem 44 (Cheeger inequality) If graph is not complete, we have h G λ h G and λ h G λ The complete graphs are much easier to understand, so we only consider the incomplete graphs Exercise 45 Find graphs G, G for which h G = c λ and h G = c λ Lemma 46 h G λ Proof Use the Rayleigh quotient, we have λ = inf f(x)dx=0 h G is achieved by S, h G = by Then λ R(g) = x y (f(x) f(y)) x f R(g) with g(x)d x = 0 (x)d x S Vol(S) with Vol(S) Vol(G) Choose g defined, if v S, Vol(S) g(v) = Vol(S c, if v / S ) ( + ) S VolS VolS c ( VolS ) VolS + ( VolS ) VolS c c = ( VolS + VolS c ) S = S VolS VolS c h G Theorem 47 λ h G Proof Suppose λ = R(f) = u v (f(u) f(v)) u f (u)d u

13 for some f Order the vertices crossing f by f(u ) f(u ) f(u n ) Let C i = {{v j, v k } j i < k n} Define C i α := α f = min i n min{ j i d j, j>i d j} h G by the definition of h G It suffices to show that λ α Now, we define V + = {v : f(v) 0} and V = {v : f(v) < 0} WLOG, we may assume f(v)<0 d v f(y) 0 d u (if not, changing sign) Define g(x) = { f(x), if x V +, 0, otherwise Then we have λ f(u)d u = (f(u) f(v)) v,v u The above equality holds since if we set φ = D f, then λ φ = Lφ, then and λ f(x)(d x ) = D (D A)D (D f(x)) = (f(x) f(y)) d x y Lg, g λ = inf g φ 0 g, g = inf f, Lf f(x)dx=0 f, Df Therefore, u V λ = + f(u) v u (f(u) f(v)) u V + f (u)d u u v (g(u) g(v)) u g (u)d u u v u v (g(u) + g(v)) (g(u) + g(v)) ( u v (g (u) g (v)) ( u g (u)d u ) ( i g (u i ) g (u i+ ) C i ) ( u g (u)d u ) ( (g (v i ) g (v i+ )) h G ( u g (u)d u ) h G ( i g(v i)( j i d j j i d j) ( u g (u)d u ) = h G, where we used the summation by parts in the last inequality Exercise 48 Prove that λ h G by using (a + b) = (a + b ) (a b) Remark 49 Cheeger s inequality can be rewritten as h G λ α h G 5 Another characterization of h G Let G = (V, E) be a graph with V = n and L = I D AD be harmonic 0 = λ λ λ n, then we have 0 < λ < Recall that we have 3

14 defined the Cheeger s constant h G = e(s, S c ) min S V VolS, VolS where e(s, S c ) = S in the previous section In fact, the Cheeger s inequality is h G λ > h G and λ α h G We have the following theorem 5 Another characterization of Cheeger s constant Theorem 5 The Cheeger s constant h G = inf sup x y f(x) f(y) f c x V f(x) c d (:= h ) x Proof First, we claim that h G h It is enough to show that holds for all f : V R Choose a constant c 0 to be the median of f, which means either d x d x, or x,f(x)<c x,f(x) c d x > x,f(x) c x,f(x)>c which was obtained by the ideas of VolS (cut f from the middle) Now, we put g = f c and for all σ > 0, we consider d x d x Let Thus, we have f(x) f(y) = x y x,g(x)<σ x,g(x)>σ d x, g(σ) = {{x, y} E g(x) σ g(y)} h G ˆ ˆ 0 which implies h h G g(σ)dσ g(σ) = ( x,g(x)<σ d x ˆ 0 h G ( ( d x )dσ + g(x)<σ = h G g(x) d x = h G x V x,g(x)<σ ˆ x V 0 x,g(x)<σ ˆ 0 d x )dσ + ( g(x)>σ f(x) c d x, d x )dσ d x g(σ) ( x,g(x)>σ d x x,g(x)>σ d x )dσ 4

15 For h G h, we suppose that h G = X for some X V, with VolX VolX VolX c Define {, if v X ϕ(v) = 0, otherwise, then we have sup c x y ϕ(x) ϕ(y) x ϕ(x) c d x which proves the theorem = sup c X c VolX + + c VolX c = X VolX = h G, Now, we give another proof for λ h G should be proved more than one method) (important mathematical theorems 5 A second proof of λ h G Proof Suppose f achieves λ, which means λ = R(f) Then λ = R(f) = sup c x y (f(x) f(y)) x y (f(x) f(y)) x (f(x) c) d x x (f(x) c 0), d x where c 0 is the median of f If we set g(x) = f(x) c 0, then λ x y (g(x) g(y)) x g (x)d x ( x y g (x) g (y)) ( x g (x)d x ) h Remark 5 Note that L = D A, x y (g(x) + g(y)) x y (g(x) + g(y)) f, Lf = x f(x) y (f(x) f(y)) = x y(f(x) f(y)), since x f(x)(f(x) f(y)) + y f(y)(f(y) f(x)) Exercise 53 Prove that h G inf f, f(x)d x=0 53 More properties f(x) f(y) x f(x) d x h G x y Let G = (V, E) be a graph as before L = I D AD = D (D A)D = D LD For edge expansion, we have h(s) = S L has eigenvalues λ 0 5

16 λ λ n and its eigenvectors are ϕ 0, ϕ,, ϕ n Let f 0, f,, f n be combinatorial eigenvectors, where f i = ϕ i D λ = inf g ϕ 0 g, Lg g, g = If f(x)d x = 0, we have inf f, Lf f(x)dx=0 f, f = f, Lf λ f, Df, inf f(x)dx=0 f, Af ( λ ) f, Df, f, Af max i 0 λ i f, Df x y (f(x) f(y)) x f (x)d x Let g G := min{ VolδS VolS VolS }, moreover, if the graph is regular, we can consider the quantity δs S Theorem 54 VolδS VolS (σ + ɛ), where σ = = λ and ɛ = Example 55 (Ramanujan k-regular) σ k k Vol(S δs) and VolδS VolS k +lower order term 4 Proof Consider f = χ S + σχ δs, where χ A is a characteristic function on A u Let g = f cχ V with c = f(u)d u, g(x)d x = 0 Note that g (x)d x = (f(x) c) d x = f (x)d x c Then f, Af = f, (D (D A))f = g + ci, (D L)(g + ci) = g, Ag + c ( λ ) g, Dg + c ( λ ) f, Df + λ c (σ + ( σ)ɛ) f, Df since c = ( u S δs f(u)d u) On the other hand, u f (u) Vol(S δs) ɛ f, Df f, Af = χ S + χ δs, A(χ S + χ δs ) = e(s, S) + σe(s, δs) + σ e(δs, δs) ( σ)e(s, S) + σe(s, S δs) = ( σ)e(s, S) + σvols σvols, where e(s, T ) = #{(u, v) u S, v T, u v} and e(s, S δs) = VolS Thus, we have (σ + ( σ)ɛ) f, Df (σ + ( σ)ɛ)(vols + σ VolδS) and VolδS VolS σ ɛ σ (σ + ɛ) (σ + ɛ) if σ, λ 6

17 If λ <, we have VolδS VolS S + (δs) VolS h(s) + h(s δs) Exercise 56 h(s) λ ( VolS ) λ ( ɛ) Theorem 57 We have where σ = max i 0 λ i 54 Diameter of graphs VolδX VolX σ, σ + VolX VolX c Consider d(u, v) =length of the shortest path joining u and v, then we have diamg = max u,v d(u, v) and d(x, Y ) = min{d(u, v) u X, v Y } Theorem 58 When X Y c, d(x, Y ) Example 59 (Ramanujan k-regular) diamg log(n ) log λ log VolXc VolY c VolX VolY log λn +λ λ n λ log n log k = log n log k Consider X, Y = X c \δx, we have 0 = D Y, (I L)D x VolX VolY σ αi βi, i 0 i 0 where D Y = α i φ i and D x = β i φ i and α 0 β 0 = will imply that σ VolX c VolY c VolX VolY, and VolX VolY This σ ( VolX)(VolX + VolδX) VolX ( VolX VolδX) and VolδX(VolX + σ ( VolX)) = VolX( VolX)( σ )) 6 Matchings and applications Definition 6 A set edges M E(G) is called matching if e, e M, e e = 7

18 6 Perfect matching Definition 6 A matching is perfect if M = V (G) in G Sometimes we can think it as a permutation σ = For X V (G), N G (X) = {y : y x for x X}, ie, covers all vertices ( ) 3 3 Theorem 63 (Hall s theorem) A bipartite graph G = (A B, E) with A = B has a perfect matching if and only if for any nonempty X A, N G (X) X or δ(x) X Proof If G has a perfect matching, it is easy to see that N G (X) X For the other direction, it suffices to show that matching covering A We use the induction on A Assume that ( ) holds for A < n and we let G have A = n Case If N G (X) > X for all nonempty X A Pick an edge {a, b} E(G), delete a, b to obtain H For = X A\{a}, N H (X) N G (X) ( X + ) = X So H has matched covering A\{a} Now, we add {a, b} back to get matching covering A in G Case Exists X A with N(X) = X By induction, matching covering X in H For Y A\X, we have N H (X) + N H (Y ) N G (X Y ) X Y = X + Y Since N H (X) = X, so N H (Y ) Y, which implies matching covering A\X in K Theorem 64 (Tutte s theorem) Let q(g) denote the number of connected components of odd order G has perfect matching if and only if for any X V (G), q(g\x) X, where G\X is a subgraph induced by V \X Proof (Sketch) To see, note that each connected component of odd order sends at least one edge to a vertex of X Each vertex of X only receives at most one edge from each component We leave as an exercise 8

19 6 Doubly stochastic matrices / magic squares Definition 65 A square matrix is doubly stochastic if nonnegative entries with row/column sums are all Example 66 P = D A of random walks on regular graph Definition 67 A magic square of weight d is a square matrix, nonnegative integers with row/column sums are all d {, if j = σ(i), Definition 68 An n n permutation matrix T σ, defined by T ij = 0, otherwise, where σ is a permutation of {,,, n} Theorem 69 (Birkhoff - Von Neumann) Every magic square of weight d is the sum of d permutation matrices Proof Let T be an n n magic square of weight d Define a bipartite G = (A B, E), where i A, j B, i j if T ij > 0 Consider X A, i X T ij = X d, so j N(X) T ij X d But j N(X) T ij = N(X) d, so there is a perfect matching by Hall s theorem, which we denote by σ Let T σ be the associated permutation matrix and T T σ is a magic square of weight d Finish with the induction on d 63 Applications: Two-sided matching problems Ordinal preferences: Find the perfect matching that is optimal with respect to preferences Definition 60 A matching is stable if and only if the following doesn t hold: r is assigned to s, but prefers r j s assignment of s j, or s j prefers r over than r j 9

20 This is related to the Gale-Shapley Algorithm Definition 6 A matching r-pareto efficient if no pair r i, r j switch assignments and both receive prefered assignment ( ) r r Here r 3 is only a stable matching not r-pareto efficient s s s 3 0